I'm using a complex analysis book that defines differentiation from an open set contained in $\mathbb R^2$ in the following manner:

Definition: Let $f:U\to\mathbb R$ continuous, where $U\subset \mathbb R^2$ is an open set. We say that $f$ is differentiable in $(x,y)\in U$, if there exists $a,b\in \mathbb R$ such that
$$\lim_{(h,k)\to (0,0)} \frac {f(x+h,y+k)-f(x,y)-ah-bh}{\sqrt {h^2+k^2}}=0.$$

1 Answer
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When we define differentiability at a certain point for functions $f : \mathbb{R} \to \mathbb{R}$ one of our goals is to assure that the function has a tangent line to the graph at the point. Indeed, the tangent line is the best linear approximation to the function near the point of tangency.

When we try to extend this notion to functions $F: \mathbb{R}^n \to \mathbb{R}^m$ we use this notion of best linear approximation to the values of the function near a point. The usual definition is (as given by Spivak on Calculus on Manifolds):

Definition: Let $F: \mathbb{R}^n \to \mathbb{R}^m$, we say that $F$ is differentiable at $a\in\mathbb{R}^n$ if there's a linear transformation $\lambda:\mathbb{R}^n \to \mathbb{R}^m$ such that

If that happens we denote $\lambda = DF(a)$ and call this linear transformation the Total Derivative of $F$ at $a$. This means that the error when approximating $F$ by $\lambda$ near $a$ goes to zero faster than the distance from $a$ to $a+h$: in other words, if you go to a neighbouring point that's close enough the function will be well approximated by the values of $\lambda$ which is a linear function by hypothesis.

The point here is - the basic idea of differentiability is this idea: approximate some function by a linear function in the neighborhood of a point. Look that this definition extends well the one dimensional case where we approximate the graph by the tangent line. Altough sometimes the way we write the definitions seems strange and complicated the ideas behind are very clear and simple.

You've presented a particular case, and it's easy to show that this definition you gave is the same definition I've presented if $n = 2$ and $m = 1$ (a good exercise is to show this). This obvious is because you can proove such a thing from the definition above and from the definition of partial derivative. The partial derivative is defined when we try to understand how a function change if we move in a particular direction. I assume that you already know the definition of partial derivative.

Although your question is: "why it's obvious that the total derivative can be expressed in terms of the partials" I think that with this information you can try to get to the answer yourself. In my opinion one of the most important parts of learning math is trying to get to the results yourself, sometimes only when you give it a try you can see what's behind.

Try it! Use this definition and the definition of partial derivatives and try to show that in the case you've presented it's true that $a = D_1f(x,y)$ and $b=D_2f(x,y)$ where $D_1$ and $D_2$ denotes the partials with respect to $x$ and $y$ respectively. If you do it try to extend to the general case.

If you can't, feel free to ask more information. Also, look at Spivak's Calculus on Manifolds, it's a very good book to learn those things.

Can I make a substitution k by 0? in order to get $a = D_1f(x,y)$
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user42912Dec 1 '12 at 13:53

If your goal is just to show that the matrix of the total derivative is the matrix of partials then perhaps the most straightforward way is that presented in Apostol's Calulus volume 2. It just uses this definition of differentiability written in another way and the definition of directional derivatives. Go ahead, try Apostol's book and if you need more information ask again.
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user1620696Dec 2 '12 at 4:15